Chemistry

Chemistry

Problem

Solid CaF2\text{CaF}_2 ( Ksp=3.9×10−11K_{sp}=3.9\times10^{-11} ) is added to 0.010 M0.010\,\text{M} HCl (pH\,2.0). Given that KaK_a for HF is 6.8×10−46.8\times10^{-4}, what is the approximate molar solubility of CaF2\text{CaF}_2 under these conditions? a) 8.5×10−5 M8.5\times10^{-5}\,\text{M} b) 1.3×10−3 M1.3\times10^{-3}\,\text{M} c) 2.2×10−4 M2.2\times10^{-4}\,\text{M} d) 2.7×10−2 M2.7\times10^{-2}\,\text{M}

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